Re: Ask for explanation -- possibly repeated

From: Pierre Weis (Pierre.Weis@inria.fr)
Date: Mon Dec 13 1999 - 17:21:41 MET


From: Pierre Weis <Pierre.Weis@inria.fr>
Message-Id: <199912131621.RAA23536@pauillac.inria.fr>
Subject: Re: Ask for explanation -- possibly repeated
To: Benoit.de-Boursetty@polytechnique.org
Date: Mon, 13 Dec 1999 17:21:41 +0100 (MET)
In-Reply-To: <Pine.GSO.4.02.9912130055340.26575-100000@young.enst.fr> from "Benoit de Boursetty" at Dec 13, 99 01:25:14 am

> > As usual: hide the side effect into the lazy side of the language that
> > is usually considered as pure and applicative, add a bit of
> > eta-expansion if your language does not know lazy evaluation properly,
> > and you get a simple workaround:
> >
> > let translate_tree t =
> > let rec aux father (Classical_node (data, sons)) =
> > let rec this_node = lazy (Node (father, data,
> > List.map
> > (aux (Some (Lazy.force this_node)))
> > sons))
> > in Lazy.force this_node
> > in aux None t;;
> > val translate_tree : 'a classical_tree -> 'a tree = <fun>
>
> Erm... "eta-expansion" is not that usual to me but... well, with all my
> respect, your solution doesn't seem to work.
> (unless it works with version 2.04, I'm still using O'CaML 2.02)
>
> # let essai = Classical_node ("Father", [Classical_node ("Son", [])]);;
> val essai : string classical_tree =
> Classical_node ("Father", [Classical_node ("Son", [])])
>
> # translate_tree essai;;
> Stack overflow during evaluation (looping recursion?).

Oups! Never forget that this is the usual result with (uncautious)
lazy evaluation :) (That's why it is so difficult).

So, the evaluation was not lazy enough, since the ``this_node'' value
was needed to compute the list of its sons.
Let me have a second try with a true lazy father field:

type 'a tree = Node of ('a tree Lazy.t) option * 'a * 'a tree list
;;

let translate_tree t =
    let rec aux father (Classical_node (data, sons)) =
      let rec this_node = lazy (Node (father, data,
                                List.map
                                  (aux (Some this_node))
                                  sons))
      in Lazy.force this_node
    in aux None t;;

Now we get:

# translate_tree essai;;
- : string tree = ...

Evidently this is not completely satisfactory, since the list of sons
are not computed in a lazy maner (this list should be a lazy
list). Hence the program is still too eager (not lazy enough), since
consider the following ``classical'' tree with recursive sharing:

# let rec bug =
 Classical_node
  (1, [Classical_node (2,[bug]); Classical_node (3, [bug; bug])]);;
val bug : int classical_tree =
...

# translate_tree bug;;
Stack overflow during evaluation (looping recursion?).

> Ah ha. So are mutable values a must even outside the Lazy module? Or does
> your eta-expansion (I don't even know what it means) need a lift... :-)

Here the eta-expansion is just a technical way to say that any
functional value f is equal to the expression
(function x -> f x).

Eta-expansion is used in the example to get a fully polymorphic
definition of the translate_tree function: we write
function x -> body x, instead of body, to expose to the typechecker
the fact that we are defining a function. Hence, we write the
(polymorphic) definition

let translate_tree t =
    ...
    in aux None t;;

instead of the monomorphic equivalent version you originally proposed

let translate_tree =
    ...
    in aux No

(More information about this in the FAQ of the language
http://pauillac.inria.fr/caml/FAQ/)

> Ah ha. So are mutable values a must even outside the Lazy module? Or does

Evidently. So is also lazy evaluation, the appropriateness of those
features depends on the problem you're trying to solve! For instance, if
you can figure out a way to solve the ``translate_tree bug'' example using
lazy evaluation, you certainly will not try to make it with mutable
values handled by hand: presumably you would say that lazy evaluation
is a must, at least inside this kind of program!

Pierre Weis

INRIA, Projet Cristal, Pierre.Weis@inria.fr, http://cristal.inria.fr/~weis/



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